Using some of these ideas, and trying to formulate my own answer based off of what I actually know, I got $lim_{x \rightarrow -N} (x+N) \Gamma(x)$ = $lim_{x \rightarrow -N} (x+N) \frac{\Gamma(x+N)}{(x)_{N}}$ = $lim_{x \rightarrow -N} \frac {\Gamma(x+N+1)}{x(x+1)(x+2)...(x+N+1)}$ =$ \frac{\Gamma(1)}{(-1)[N(N-1)(N-2)...1]}$ =$\frac{1}{(-1)(N!)}$ which is still wrong I think.
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renagade629Oct 7 '12 at 6:10

I can follow your work easiest since this formula is in my reference sheet. I don't understand the purpose of making the y = x+n substitution. Is it necessary? Can it be done without it? For $\Gamma (1+n-y)$ it disappeared from your second line when you transitioned to the third line to $\Gamma (1+n)$. Wasn't the $sin(\pi y - \pi n) = (-1)^n sin(\pi y)$ in the denominator? How did the $(-1)^n$ get moved to the numerator?
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renagade629Oct 7 '12 at 5:47

@renagade629 There is no need to introduce $y$. I thought it would be easier. Anyway, moving $(-1)^n$ from the denominator to the numerator is done as follows: $$ \frac{1}{(-1)^n} = \frac{(-1)^n}{(-1)^{2n}}$$ but for integer $n$, $2n$ is even, and thus $(-1)^{2n} = 1$.
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SashaOct 7 '12 at 12:41

Recall that $\Gamma(x)$ has simple poles at the nonpositive integers, with residue at $x=-N$ equal to $l=\frac{(-1)^N}{N!}.$ Thus, near $x=-N,$ we can express $\Gamma(x)=\frac{l}{x+N}+h(x)$ with $h(x)$ holomorphic. We obtain

as desired. Of course, the real crux of the question is to prove that the residue of the gamma function at $x=-N$ is $l.$ I'll leave this to you, because it really is a nice exercise on the idea of analytic continuation.

Thanks I will look into it. I am taking an undergraduate course in differential equations and we are using the gamma function for bessel ODE and we are using this notation. So I know nothing of the terminology that you used. I just have a sheet with gamma properties.
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renagade629Oct 7 '12 at 2:48

You're welcome! The important part is that $\Gamma(x)$ can be extended to a function on the complex numbers, minus the points $x=0,-1,-2,-3,\ldots.$ However, near each of these there is a Laurent expansion (like a Taylor series except allowing negative powers) of $\Gamma(x).$ So, in some small neighbourhood of $x=-N,$ we know that $\Gamma(x)=\frac{(-1)^N}{N!(x+N)}+a_0+a_1(x+N)+a_2(x+N)^2+\cdots,$ where the $a_i$ are constants. (In my answer, $h(x)=a_0+\cdots.$) This is extremely useful, used to prove identities such as Euler's reflection mentioned in another answer.
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AndrewOct 7 '12 at 3:18